What is the integral of e^ssin(t-s) ?

The integral of e^ssin(t-s) is (e^scos(t-s))/2+(e^ssin(t-s))/2+C

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integral of e^ssin(t-s)

\int e^{s} sin (t - s) d s

\frac{e ^{s} cos ( t - s )}{2} + \frac{e ^{s} sin ( t - s )}{2} + C

Solution steps

\int e^{s} sin (t - s) d s

Apply Integration By Parts

: e^{s} cos (t - s) - \int e^{s} cos (t - s) d s

= e^{s} cos (t - s) - \int e^{s} cos (t - s) d s

Apply Integration By Parts

: - e^{s} sin (t - s) - \int - e^{s} sin (t - s) d s

= e^{s} cos (t - s) - (- e^{s} sin (t - s) - \int - e^{s} sin (t - s) d s)

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= e^{s} cos (t - s) - (- e^{s} sin (t - s) - (- \int e^{s} sin (t - s) d s))

Therefore

\int e^{s} sin (t - s) d s = e^{s} cos (t - s) - (- e^{s} sin (t - s) - (- \int e^{s} sin (t - s) d s))

Isolate

\int e^{s} sin (t - s) d s

= \frac{e ^{s} cos ( t - s )}{2} + \frac{e ^{s} sin ( t - s )}{2}

Add a constant to the solution

= \frac{e ^{s} cos ( t - s )}{2} + \frac{e ^{s} sin ( t - s )}{2} + C

\frac{d}{d x} (3 x^{2} 25 - x^{2})

\int \frac{8 x ^{2} + 8 x + 2}{( 4 x ^{2} + 1 ) ^{2}} d x

\int_{1}^{2} \frac{1}{x ^{3} + x} d x

f (x) = x^{2} - x + 2, at x = 2

\int_{0}^{1} \frac{e ^{x}}{1 + e ^{2 x}} d x

What is the integral of e^ssin(t-s) ?
The integral of e^ssin(t-s) is (e^scos(t-s))/2+(e^ssin(t-s))/2+C